Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 2}{x + 1} = \dfrac{4x + 3}{x + 1}$
Multiply both sides by $x + 1$ $ \dfrac{x^2 - 2}{x + 1} (x + 1) = \dfrac{4x + 3}{x + 1} (x + 1)$ $ x^2 - 2 = 4x + 3$ Subtract $4x + 3$ from both sides: $ x^2 - 2 - (4x + 3) = 4x + 3 - (4x + 3)$ $ x^2 - 2 - 4x - 3 = 0$ $ x^2 - 5 - 4x = 0$ Factor the expression: $ (x + 1)(x - 5) = 0$ Therefore $x = -1$ or $x = 5$ At $x = -1$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -1$, it is an extraneous solution.